Calculus I

NU Social Science Math Camp

What have we learned so far?

Foundational ideas behind what you will have to deal with during your graduate program:

  • Reading about methods \(\rightarrow\) Notation, sets, functions
  • Working with data frames \(\rightarrow\) Matrix operations

Up next

Dealing with continuity (which implies infinity) \(\rightarrow\) Calculus

Two new operators:

  1. Derivatives (Today)

  2. Integrals (Friday)

Calculus

  • This is often where people experience the most difficulty

  • We are now operating with whole functions!

  • New operators have new, different rules

  • We will stick together!

Derivatives as a slope

  • How do we find the slope of this line?

  • \(m = \frac{f(x_2) - f(x_1)}{x_2-x_1} = \frac{rise}{run}\)

  • Let’s try it out!

Derivatives as rate of change

  • Positive derivative \(\rightarrow\) Function is increasing

  • Negative derivative \(\rightarrow\) Function is decreasing

  • Zeroes \(\rightarrow\) Max or min (more soon!)

Now try this one!

\(y = x^2\)

Why doesn’t it work?

  • We need an operation to calculate the slope or rate of change for (almost) any kind of continuous function

  • Enter derivatives

Generalize slope formula

\[ m = \frac{f(x_1+ \Delta x) - f(x_1)}{\Delta x} \]

Derivatives

When \(\Delta x\) is small enough

\[ \lim_{\Delta x\to0} \frac{f(x_1+ \Delta x) - f(x_1)}{\Delta x} = \frac{d}{dx} f(x) = f'(x) \]

  • Read \(\frac{d}{dx} f(x)\) as “the derivative of \(f\) of \(x\) with respect to \(x\)”

  • Read \(f'(x)\) as “\(f\) prime \(x\)” for a shorthand

How to compute a derivative?

  • Plug in many values into \(\frac{rise}{run}\)
  • Use properties of limits
  • Solve analytically
  • Software

How to compute a derivative?

  • Plug in many values into \(\frac{rise}{run}\)
  • Use properties of limits
  • Solve analytically
  • Software

Using limits

\[ \lim_{\Delta x\to0} \frac{f(x_1+ \Delta x) - f(x_1)}{\Delta x} \]

Using limits

\[ \lim_{h\to0} \frac{f(x_1+ h) - f(x_1)}{h} \]

Back to \(x^2\)

\[ \lim_{h \to0} \frac{f(x+ h)^2 - x^2}{h} \]

Using limits

\[ \lim_{h\to0} \frac{f(x_1+ h) - f(x_1)}{h} \]

Back to \(x^2\)

\[ \lim_{h\to0} \frac{f(x+ h)^2 - x^2}{h} = \lim_{h\to0} \frac{x^2 + 2xh + 2h^2 - x^2}{h} \]

Using limits

\[ \lim_{h\to0} \frac{f(x_1+ h) - f(x_1)}{h} \]

Back to \(x^2\)

\[ \lim_{h\to0} \frac{f(x+ h)^2 - x^2}{h} = \lim_{h\to0} \frac{x^2 + 2xh + 2h^2 - x^2}{h} = \lim_{h \to 0} \frac{2xh + 2h^2}{h} \]

Using limits

\[ \lim_{h\to0} \frac{f(x_1+ h) - f(x_1)}{h} \]

Back to \(x^2\)

\[ \lim_{h\to0} \frac{f(x+ h)^2 - x^2}{h} = \lim_{h\to0} \frac{x^2 + 2xh + 2h^2 - x^2}{h} = \lim_{h \to 0} \frac{2xh + 2h^2}{h} \] \[ = \lim_{h \to 0} \frac{(2x+2h) h}{h} \]

Using limits

\[ \lim_{h\to0} \frac{f(x_1+ h) - f(x_1)}{h} \]

Back to \(x^2\)

\[ \lim_{h\to0} \frac{f(x+ h)^2 - x^2}{h} = \lim_{h\to0} \frac{x^2 + 2xh + 2h^2 - x^2}{h} = \lim_{h \to 0} \frac{2xh + 2h^2}{h} \] \[ = \lim_{h \to 0} \frac{(2x+2h) h}{h} = \lim_{h \to 0} 2x+2h \]

Using limits

\[ \lim_{h\to0} \frac{f(x_1+ h) - f(x_1)}{h} \]

Back to \(x^2\)

\[ \lim_{h\to0} \frac{f(x+ h)^2 - x^2}{h} = \lim_{h\to0} \frac{x^2 + 2xh + 2h^2 - x^2}{h} = \lim_{h \to 0} \frac{2xh + 2h^2}{h} \] \[ = \lim_{h \to 0} \frac{(2x+2h) h}{h} = \lim_{h \to 0} 2x+2h \] \[ =2x+0 \]

Using limits

\[ \lim_{h\to0} \frac{f(x_1+ h) - f(x_1)}{h} \]

Back to \(x^2\)

\[ \lim_{h\to0} \frac{f(x+ h)^2 - x^2}{h} = \lim_{h\to0} \frac{x^2 + 2xh + 2h^2 - x^2}{h} = \lim_{h \to 0} \frac{2xh + 2h^2}{h} \] \[ = \lim_{h \to 0} \frac{(2x+2h) h}{h} = \lim_{h \to 0} 2x+2h \] \[ =2x+0 = 2x \]

How to compute a derivative?

  • Plug in many values into \(\frac{rise}{run}\)
  • Use properties of limits
  • Solve analytically
  • Software

How to compute a derivative?

  • Plug in many values into \(\frac{rise}{run}\)
  • Use properties of limits
  • Solve analytically
  • Software

Rules of differentiation

Differentation: “Taking the derivative”

Rules of differentiation

Constant rule: \((c)' = 0\).

\(y = 2\)

Rules of differentiation

Coefficient rule: \((c \cdot f(x))' = c \cdot f'(x)\)

Rules of differentiation

Sum/difference rule: \((f(x) \pm g(x))' = f'(x) \pm g'(x)\).

Example: \(f(x) = 3x + 7\)

\[ f'(x) = (3x + 7)' \]

\[ = (3x)' + (7)' \]

\[ = 3 + 0 \]

\[ = 3 \]

Rules of differentiation

Power rule: \((x^n)'=nx^{(n-1)}\)

\[ f(x) = x^2 \]

\[ f'(x) = nx^{(n-1)} \]

\[ = 2x^{(2-1)} \]

\[ = 2x \]

  • Rate of change depends on values of \(x\)
  • “… at a given point”

Practice

Let’s try out \(\frac{d}{dx}4x^3\) and \(\frac{d}{dx}(x^2 + 2x)\) on the board.

More practice

Use the differentiation rules we have covered so far to calculate the derivatives of \(y\) with respect to \(x\) of as many as these as you can.

  1. \(y = 2x^2 + 10\)
  2. \(y = 5x^4 - \frac{2}{3}x^3\)
  3. \(y = 9 \sqrt x\)
  4. \(y = \frac{4}{x^2}\)
  5. \(y = \frac{2w}{5}\)

Rules of differentiation

Exponent and logarithm rules:

\[ \begin{aligned} (c^x)' &= c^x \cdot ln(c), & \forall x>0 \\ \\ (log_a(x))' &= \frac{1}{x \cdot ln(a)}, & \forall x>0 \\ \end{aligned} \]

Rules of differentiation

Exponent and logarithm rules:

Euler’s number (\(e\)) has interesting properties when it comes to derivatives.

\[ \begin{aligned} (e^x)' &= e^x \\ \\ (ln(x))' &= \frac{1}{x}, & \forall x>0 \end{aligned} \]

More more practice

Compute the following:

  1. \(\frac{d}{dx}(10e^x)\)
  2. \(\frac{d}{dx}(ln(x) - \frac{e^2}{3})\)

Advanced rules

Product rule: \((f(x)g(x))'=f'(x)g(x) + g'(x)f(x)\)

\[ [(3x+4)(x+2)]' \]

\[ f(x) = (3x+4), g(x) = (x+2); f'(x) = 3, g'(x) = 1 \]

\[ \color{blue}{f'(x)}\color{yellow}{g(x)} + \color{purple}{g'(x)}\color{green}{f(x)} \]

\[ = \color{blue}3 \color{yellow}{(x+2)} + \color{purple}{1} \color{green}{(3x+4)} \]

\[ = 3x + 6 + 3x+ 4 \]

\[ = 6x + 10 \]

Advanced rules

Quotient rule: \(\displaystyle(\frac{f(x)}{g(x)})' = \frac{f'(x)g(x) + g'(x)f(x)}{[g(x)]^2}\)

\[ \frac{3x^2}{x+2} \]

Advanced rules

Quotient rule: \(\displaystyle(\frac{f(x)}{g(x)})' = \frac{f'(x)g(x) + g'(x)f(x)}{[g(x)]^2}\)

\[ \frac{3x^2}{x+2} \Rightarrow f(x) = 3x^2, g(x) = x + 2; f'(x) = 6x, g'(x) = 1 \]

\[ \frac{\color{blue}{f'(x)}\color{yellow}{g(x)} + \color{purple}{g'(x)}\color{green}{f(x)}}{[\color{yellow}{g(x)}]^2} = \frac{\color{blue}{6x}\color{yellow}{(x+2)} - \color{purple}1 \color{green}{(3x^2)}}{\color{yellow}{(x+2)}^2} \]

\[ = \frac{6x^2 + 12x - 3x2}{(x+2)^2} = \frac{3x^2+12x}{(x+2)^2} \]

Advanced rules

Chain rule: \((f(g(x))' = f'(g(x)) \cdot g'(x)\)

\[ f(g(x)) = (x+3)^3 \]

Advanced rules

Chain rule: \([f(g(x))]' = f'(g(x)) \cdot g'(x)\)

\[ f(g(x)) = (x+3)^3 \Rightarrow f(g(x)) = g(x)^3, g(x) = (x+3) \]

\[ f'(x) = 3x^2 \Rightarrow f'(g(x)) = 3 (x+3)^2, g'(x) = 1 \]

Substitute

\[ [f(g(x))]' = 3(x+3)^2 \cdot 1 \]

More more more practice

  1. \(\frac{d}{dx}x^3 \cdot x\)
  2. \(\frac{d}{dx}e^x \cdot x^2\)
  3. \(\frac{d}{dx}(3x^4-8)^2\)

Higher-order derivatives

  • \(f'(x)\): First derivative
  • \(f''(x)\): Second derivative
  • \(f'''(x)\)?

If we were driving a car:

  • \(f(x)\) = distance traveled at time \(x\)
  • \(f'(x)\) = speed at time \(x\)
  • \(f''(x)\) = acceleration at time \(x\)

To get \(f''(x)\) just take the derivative of the output of \(f'(x)\)

Partial derivative

What happens when we have more than one variable? \(f(x,z)\)

Partial derivative

\[ \frac{\partial}{\partial_x}f(x,z) = \frac{\partial_y}{\partial_x} = \partial_x f \]

Just treat other variables as constants and calculate the derivative of the target variable (usually \(x\))

Example

\[ \begin{aligned} y = f(x,z) &= xz \\ \frac{\partial_y}{\partial_x} &= z \end{aligned} \]

Let’s try \(\displaystyle\frac{\partial (x^2y+xy^2-x)}{\partial x}\) and \(\displaystyle\frac{\partial (x^2y+xy^2-x)}{\partial y}\) together

Differentiability of functions

  • Not all functions are differentiable
  • We need continuous and reasonably smooth functions
  • Informally: Can draw them without lifting the pen
  • Formally: A function is continuous at a point \(a\) if: \(\lim_{x \to a} f(x)=f(a)\)
  • Differentiable \(\to\) continuous
  • Continuous \(\not\to\) differentiable

How to compute a derivative?

  • Plug in many values into \(\frac{rise}{run}\)
  • Use properties of limits
  • Solve analytically
  • Software

How to compute a derivative?

  • Plug in many values into \(\frac{rise}{run}\)
  • Use properties of limits
  • Solve analytically
  • Software

Using computers

  • Symbolic differentiation: Do the math for you (WolframAlpha or Mathematica)
  • Numerical differentiation: Evaluate the function at many values (optim() in R)
  • Automatic differentiation: Break up functions into elementary operations and use chain rule (torch package in R, various packages in Python)

Optimization

  • Find minimum or maximum values

  • Valleys are minima

  • Peaks are maxima

  • Both are extrema

  • Extrama can be local or global

First order condition

  • Minimum: Derivative is negative before and positive after

  • Maximum: Derivative is positive before and negative after

  • Extrema are critical points

  • First order condition helps us find them

\[FOC: f'(x)=0\]

Second order condition

FOC finds critical points, but can’t tell if minimum or maximum

Second derivative test

  • Start by identifying \(f''(x)\)

  • Substitute in the stationary points \((x^*)\) identified from the FOC

  • \(f''(x^*) > 0\) we have a local minimum

  • \(f''(x^*) < 0\) we have a local maximum

  • \(f''(x^*) = 0\) we (may) have an inflection point - need to calculate higher-order derivatives

More more more more practice

Obtain FOC and SOC for \(\displaystyle y=\frac{1}{2} x^3 + 3 x^2 - 2\)

Local vs. global extrema

Extreme value theorem:

  • If a real-valued function is continuous on a closed and bounded (finite) interval, the function must have a global minimum and a global maximum on that interval at least once

  • Either at the boundaries or at one of the local extrema

  • To find the minimum/maximum on some interval, compare the local min/max to the value of the function at the interval’s endpoints. So, if the interval is \((-\infty, +\infty)\), check the function’s limits as it approaches \(-\infty\) and \(+\infty\).